Sent from my iPhone
On Sep 25, 2008, at 10:39 AM, "Miguel A. Quintans" <[EMAIL PROTECTED]
> wrote:
Hello
The result of an expression using pre-decrement or pre-increment
such as:
y = x * n * --n;
Try turning on warnings. That is -Wsquence-points. The above is
specified behavior as there are squence points between the increment
of n and the other access of n.
gives different results in a few cases where x is placed before or
after the rest, or when its value is 1 or not. Please run the
program attached where a comment indicates what we think works
wrong. We obtain:
1.- 16
2.- 16
3.- 16
4.- 40
5.- 20
6.- 16
7.- 40
8.- 32
1.- 36
2.- 36
3.- 36
4.- 60
5.- 30
6.- 36
7.- 60
8.- 72
In contrast, the results we get running it in a AIX Computer with
the native compiler (IBM XL C/C++ Enterprise Edition for AIX, V9.0
Version: 09.00.0000.0000) is what we expected:
1.- 20
2.- 20
3.- 20
4.- 40
5.- 20
6.- 20
7.- 40
8.- 40
1.- 30
2.- 30
3.- 30
4.- 60
5.- 30
6.- 30
7.- 60
8.- 60
We have experienced this issue in several versions of the gcc
compiler (the ones distributed under fedora cores 6, 7, 8, knoppix
5.0.1 and also in the windows Dev-c++ 4.9.9.2 package compiler)
obtaining the same results, that we think are (obviously) wrong.
We remain expectant of any comment from you.
Best regards
Miguel Ángel Quintáns & Virginia Escuder
Professors at the Universidad de Alcalá
Madrid
Spain
#include int main () { int x=2; int p=1; int n=5; int y = n * --n;
printf ("1.- %d\n",y); // wrong n=5; y = 1 * n * --n; printf ("2.- %d
\n",y); // wrong n=5; y = n * --n * 1; printf ("3.- %d\n",y); //
wrong n=5; y = 2 * n * --n; printf ("4.- %d\n",y); n=5; y = p * n *
--n; printf ("5.- %d\n",y); n=5; y = n * --n * p; printf ("6.- %d
\n",y); // wrong n=5; y = x * n * --n; printf ("7.- %d\n",y); n=5; y
= n * --n * x; printf ("8.- %d\n",y); // wrong n=5; y = n * ++n;
printf ("1.- %d\n",y); // wrong n=5; y = 1 * n * ++n; printf ("2.- %d
\n",y); // wrong n=5; y = n * ++n * 1; printf ("3.- %d\n",y); //
wrong n=5; y = 2 * n * ++n; printf ("4.- %d\n",y); n=5; y = p * n * +
+n; printf ("5.- %d\n",y); n=5; y = n * ++n * p; printf ("6.- %d
\n",y); // wrong n=5; y = x * n * ++n; printf ("7.- %d\n",y); n=5; y
= n * ++n * x; printf ("8.- %d\n",y); // wrong }
